Most people think normal body temperature is 98.6 °F. In 1992, the Journal of the American Medical Association asserted that a more accurate figure may be 98.2 °F on average, and that body temperatures had a standard deviation of 0.7 °F. Assuming this is true and body temperatures follow a normal distribution, what is the probability that a body temperature will be above 99 °F?

Options:

A) 0.1271
B) 0.5714
C) 0.2843
D) 0.8729
E) 0.7157

To find the probability of a body temperature above 99 °F, we can calculate the Z-score and use a Z-score table or calculator. The probability is approximately 0.1271.

Explanation:

To find the probability that a body temperature will be above 99 °F, we can use the Z-score formula. The formula is Z = (x - μ) / σ, where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation. Using the given information, we have a mean (μ) of 98.2 °F and a standard deviation (σ) of 0.7 °F.

Plugging the values into the formula, we get Z = (99 - 98.2) / 0.7. Calculating this, we find Z = 0.8 / 0.7 = 1.14.

Using a Z-score table or calculator, we can find that the probability of a Z-score of 1.14 or greater is approximately 0.1271. Therefore, the probability that a body temperature will be above 99 °F is approximately 0.1271.

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